Question

A fish swims to a depth of 50.00 meters in the ocean. Assuming the density of sea water is 1.0251.025 g·cm^{-3}g⋅cm −3 , calculate how much water pressure the fish is experiencing at this depth in units of kPa.

Answer 1

Answer:

The fish is experiencing a water pressure of 502.8 kPa.

Explanation:

The water pressure the fish is experiencing can be found as follows:

$P = \rho gh$  (1)

Where:

g: is the gravity = 9.81 m/s²

h: is the height (depth) = 50.0 m

ρ: is the seawater's density = 1.025 g/cm³  

By replacing the above values into equation (1) we have:

$P = \rho gh = 1.025 \frac{g}{cm^{3}}*\frac{1 kg}{1000 g}*\frac{(100cm)^{3}}{1 m^{3}}*9.81 m/s^{2}*50.0 m = 502.8 kPa$        

Therefore, the fish is experiencing a water pressure of 502.8 kPa.

I hope it helps you!